# How do we define the cotangent space as the quotient of ideals?

I am interested in the definition of the cotangent space as the quotient space of ideals. The definition goes like this:

Let $$\mathcal M$$ be a smooth manifold. $$C^\infty (\mathcal M)$$ is the ring of smooth scalar fields on $$\mathcal M$$. Let $$\mathcal I_p$$ be the greatest subring of $$C^\infty (\mathcal M)$$ where $$\phi (p)=0$$ for all $$\phi \in \mathcal I_p$$. $$\mathcal I_p$$ is an ideal. The square of this ideal is $$\mathcal I_p^2=\{\sum_{i=1}^n \phi_i \psi_i | n\in \Bbb N, \phi_i,\psi_i\in\mathcal I_p \}$$. $$\mathcal I_p$$ and $$\mathcal I_p^2$$ are vector spaces. The quotient space of $$\mathcal I_p$$ and $$\mathcal I_p^2$$ is $$\mathcal I_p/\mathcal I_p^2=\{\phi+\mathcal I_p^2|\phi\in\mathcal I_p\}$$. This quotient space is either equal to the cotangent space on $$\mathcal M$$ at $$p$$ or isomorphic to it.

An element of $$\mathcal I_p/\mathcal I_p^2$$ could look like $$\Phi=\{\phi + \psi \gamma, \phi + \eta \nu, \phi + \delta \upsilon + \alpha \beta,... \}$$. How is $$\Phi$$ interpreted as a covector such as $$\text d\phi_p$$? Is $$\Phi$$ equal to $$\text d \phi_p$$?

• Also asked here. Commented Jul 23, 2023 at 13:42

This is basically an algebraic construction, that goes as follows: suppose $$M$$ is a smooth manifold of dimension $$n$$.

$$1).\$$ For fixed $$p\in M$$, define $$m_{p}$$ $$\subseteq C^{\infty}(M)$$ to be the subalgebra of functions such that $$f(p)=0.$$

$$2).\$$ Let $$v$$ be a derivation in $$T_pM$$ and define $$m_p^2:=\{h\in C^{\infty}(M):h=fg \ \text{for some }\ f,g\in m_p\}.$$ Since $$v$$ is a derivation, it follows easily from $$1).$$ that $$h\in m_p^2\Rightarrow v(h)=0.$$

$$3).\$$ Define $$\Phi_v:m_p/m^2_p \rightarrow \mathbb R$$ by $$\Phi_{v}(\varphi+fg)=v(\varphi).$$ This map is well-defined because $$v(fg)=0$$ and it is obviously linear. It follows that every $$v\in T_pM$$ uniquely determines such a linear map $$\Phi_v$$.

$$4).\$$ Now let $$\Phi:m_p/m^2_p \rightarrow \mathbb R$$ be any linear functional. Define $$v:C^{\infty}(M)\to \mathbb R$$ by $$v(f)=\Phi\circ \pi(f-f(p))$$ where $$\pi$$ is the evident quotient map. It is tedious but routine to show that $$v$$ is a derivation.

$$5).\ 3).\$$ and $$4).\$$ combine to show that $$(m_p/m_p^2)^*\cong T_pM$$ and in particular shows that $$\dim m_p/m_p^2=\dim (m_p/m_p^2)^*=\dim T_pM=n.$$

$$6).\$$ One can also prove finite dimensionality of $$m_p/m_p^2$$ directly: take local coordinates $$(U,x)$$ about $$p\in M$$. Then $$x^i(p)=0$$ for $$1\le i\le n$$. If $$f\in m_p,$$ then $$f(x)=0$$ at $$p$$. Now, apply Hadamard's lemma to find $$g_i\in C^{\infty}(U)$$ such that $$f(x)=\sum^n_{k=1} x^kg_k(x)=xg_1(x)+\sum^n_{k=2}x^kg_k(x)$$ and since $$\sum^n_{k=2}x^kg_k(x)\in m_p^2$$ we have that $$f\sim xg_1.$$ That is, $$f(x^1,\cdots,x^n)\sim (x^1,\cdots,x^n)\cdot g_1(x^1,\cdots,x^n)$$ and the RHS of this spans a vector space of dimension $$n$$ as $$f$$ varies through $$C^{\infty}(M).$$